3.61 \(\int (f x)^m \log (c (d+e \sqrt{x})^p) \, dx\)

Optimal. Leaf size=83 \[ \frac{(f x)^{m+1} \log \left (c \left (d+e \sqrt{x}\right )^p\right )}{f (m+1)}-\frac{e p x^{3/2} (f x)^m \, _2F_1\left (1,2 m+3;2 (m+2);-\frac{e \sqrt{x}}{d}\right )}{d \left (2 m^2+5 m+3\right )} \]

[Out]

-((e*p*x^(3/2)*(f*x)^m*Hypergeometric2F1[1, 3 + 2*m, 2*(2 + m), -((e*Sqrt[x])/d)])/(d*(3 + 5*m + 2*m^2))) + ((
f*x)^(1 + m)*Log[c*(d + e*Sqrt[x])^p])/(f*(1 + m))

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Rubi [A]  time = 0.0505886, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2455, 20, 341, 64} \[ \frac{(f x)^{m+1} \log \left (c \left (d+e \sqrt{x}\right )^p\right )}{f (m+1)}-\frac{e p x^{3/2} (f x)^m \, _2F_1\left (1,2 m+3;2 (m+2);-\frac{e \sqrt{x}}{d}\right )}{d \left (2 m^2+5 m+3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^m*Log[c*(d + e*Sqrt[x])^p],x]

[Out]

-((e*p*x^(3/2)*(f*x)^m*Hypergeometric2F1[1, 3 + 2*m, 2*(2 + m), -((e*Sqrt[x])/d)])/(d*(3 + 5*m + 2*m^2))) + ((
f*x)^(1 + m)*Log[c*(d + e*Sqrt[x])^p])/(f*(1 + m))

Rule 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int (f x)^m \log \left (c \left (d+e \sqrt{x}\right )^p\right ) \, dx &=\frac{(f x)^{1+m} \log \left (c \left (d+e \sqrt{x}\right )^p\right )}{f (1+m)}-\frac{(e p) \int \frac{(f x)^{1+m}}{\left (d+e \sqrt{x}\right ) \sqrt{x}} \, dx}{2 f (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+e \sqrt{x}\right )^p\right )}{f (1+m)}-\frac{\left (e p x^{-m} (f x)^m\right ) \int \frac{x^{\frac{1}{2}+m}}{d+e \sqrt{x}} \, dx}{2 (1+m)}\\ &=\frac{(f x)^{1+m} \log \left (c \left (d+e \sqrt{x}\right )^p\right )}{f (1+m)}-\frac{\left (e p x^{-m} (f x)^m\right ) \operatorname{Subst}\left (\int \frac{x^{-1+2 \left (\frac{3}{2}+m\right )}}{d+e x} \, dx,x,\sqrt{x}\right )}{1+m}\\ &=-\frac{e p x^{3/2} (f x)^m \, _2F_1\left (1,3+2 m;2 (2+m);-\frac{e \sqrt{x}}{d}\right )}{d \left (3+5 m+2 m^2\right )}+\frac{(f x)^{1+m} \log \left (c \left (d+e \sqrt{x}\right )^p\right )}{f (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0339509, size = 76, normalized size = 0.92 \[ \frac{x (f x)^m \left (d (2 m+3) \log \left (c \left (d+e \sqrt{x}\right )^p\right )-e p \sqrt{x} \, _2F_1\left (1,2 m+3;2 m+4;-\frac{e \sqrt{x}}{d}\right )\right )}{d (m+1) (2 m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^m*Log[c*(d + e*Sqrt[x])^p],x]

[Out]

(x*(f*x)^m*(-(e*p*Sqrt[x]*Hypergeometric2F1[1, 3 + 2*m, 4 + 2*m, -((e*Sqrt[x])/d)]) + d*(3 + 2*m)*Log[c*(d + e
*Sqrt[x])^p]))/(d*(1 + m)*(3 + 2*m))

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Maple [F]  time = 0.328, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m}\ln \left ( c \left ( d+e\sqrt{x} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*ln(c*(d+e*x^(1/2))^p),x)

[Out]

int((f*x)^m*ln(c*(d+e*x^(1/2))^p),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e*x^(1/2))^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (f x\right )^{m} \log \left ({\left (e \sqrt{x} + d\right )}^{p} c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e*x^(1/2))^p),x, algorithm="fricas")

[Out]

integral((f*x)^m*log((e*sqrt(x) + d)^p*c), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*ln(c*(d+e*x**(1/2))**p),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{m} \log \left ({\left (e \sqrt{x} + d\right )}^{p} c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*log(c*(d+e*x^(1/2))^p),x, algorithm="giac")

[Out]

integrate((f*x)^m*log((e*sqrt(x) + d)^p*c), x)